Trying to evaluate $\int_{0}^{\infty}\frac{\ln(1+x^3)}{1+x^3}\frac{dx}{1+x^3}$ I would like to work this out:

$$I=\large\int_{0}^{\infty}\frac{\ln(1+x^3)}{1+x^3}\frac{\mathrm dx}{1+x^3}$$

Making a sub: $u=x^3$, $dx=\frac{du}{3x^2}$
$$I=\frac{1}{3}\int_{0}^{\infty}\frac{\ln(1+u)}{u^{2/3}(1+u)^2}\mathrm du$$
Making a sub: $u=\tan^2(y)$, $du=\sec^2(y)dy$
$$I=\frac{2}{3}\int_{0}^{\pi/2}\frac{\ln\sec(y)}{\sec^2(y)\sqrt[3]{\tan^4(y)}}\mathrm dy$$
$$I=\frac{2}{3}\int_{0}^{\pi/2}\frac{\cot(y)\cos^2(y)\ln\sec(y)}{\sqrt[3]{\tan(y)}}\mathrm dy$$
I can't continue.
Maybe there is another alternative way to simplify $I$
 A: The substitution $t = (1 + x^3)^{-1}$ yields
$$ I = \frac{1}{3} \int \limits_0^1 - \ln(t) \left(\frac{t}{1-t}\right)^{2/3} \, \mathrm{d} t = f'\left(\frac{2}{3}\right) \, , $$
where
$$ f(\alpha) \equiv - \frac{1}{3} \int \limits_0^1 \frac{t^\alpha}{(1-t)^{2/3}} \, \mathrm{d} t = - \frac{1}{3} \operatorname{B} \left(\frac{1}{3}, \alpha +1 \right) = - \frac{\operatorname{\Gamma} \left(\frac{1}{3}\right)}{3} \frac{\operatorname{\Gamma}(\alpha + 1)}{\operatorname{\Gamma}\left(\alpha + \frac{4}{3} \right)} ~~~ , \, \alpha > -1 \, .$$
The differentiation under the integral sign can be justified using the dominated convergence theorem.
In terms of the digamma function $\psi$ we now have
$$ I = f'\left(\frac{2}{3}\right) = \frac{2}{9} \operatorname{\Gamma} \left(\frac{1}{3}\right) \operatorname{\Gamma} \left(\frac{2}{3}\right) \left[\operatorname{\psi} (2) - \operatorname{\psi} \left(\frac{5}{3}\right)\right] \, . $$
We can use the reflection formulas for $\Gamma$ and $\psi$ and the recurrence relation $\operatorname{\psi}(x + 1) = \operatorname{\psi}(x) + \frac{1}{x}$ to find
$$ I = \frac{2}{9} \frac{\pi}{\sin\left(\frac{\pi}{3}\right)} \left[\operatorname{\psi}(1) - \operatorname{\psi} \left(\frac{1}{3}\right) - \pi \cot\left(\frac{\pi}{3}\right) - \frac{1}{2} \right] \, .$$
The special values $\operatorname{\psi}(1) = - \gamma$ and $\operatorname{\psi} \left(\frac{1}{3}\right) = - \frac{\pi}{2 \sqrt{3}} - \frac{3}{2} \ln (3) - \gamma$ then lead to the final result
$$ I = \frac{2 \pi}{27} [\sqrt{3} (3 \ln(3) - 1) - \pi] \, .$$
A: In this solution we will not use the beta function or the digamma function.
Put
\begin{equation*}
f(s)=\int_{0}^{\infty}\dfrac{\ln(1+s^3x^3)}{(1+x^3)^2}\, \mathrm{d}x.
\end{equation*}
We want to calculate $f(1)$. However,
\begin{equation*}
f(1) = f(1)-f(0) = \int_{0}^{1}f'(s)\, \mathrm{d}s\tag{1}
\end{equation*}
and
\begin{equation*}
f'(s)=\int_{0}^{\infty}\dfrac{3s^2x^3}{(1+s^3x^3)(1+x^3)^2}\, \mathrm{d}x =[x^3=z] = \int_{0}^{\infty}\dfrac{s^2z^{1/3}}{(1+s^3z)(1+z)^2}\, \mathrm{d}z .
\end{equation*}
For $0<s<1$ we use keyhole contour integration and get
\begin{equation*}
f'(s) = \dfrac{2\pi\sqrt{3}}{9}\dfrac{2s+1}{(s^2+s+1)^2}\cdot s^2 .
\end{equation*}
Integration by parts in (1) will finally give us
\begin{equation*}
f(1) = \dfrac{2\pi}{27}\left(\sqrt{3}(3\ln 3 -1)-\pi\right).
\end{equation*}
